# How do I find the derivative of f(x) = sqrt(x+3) using first principles?

Apr 20, 2018

$f ' \left(x\right) = \frac{1}{2 \sqrt{x + 3}}$

#### Explanation:

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f \left(x\right) = \sqrt{x + 3} , f \left(x + h\right) = \sqrt{x + h + 3}$, then

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h}$

If we evaluate this right away, we get

${\lim}_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h} = \frac{\sqrt{x + 3} - \sqrt{x + 3}}{0} = \frac{0}{0}$,

so we need to simplify as this is an indeterminate form.

Multiply the entire limit by the numerator's conjugate, which is $\frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}}$. This is the same as multiplying by $1.$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{x + h + 3} - \sqrt{x + 3}}{h} \cdot \frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}}$

The numerator becomes

$\sqrt{x + h + 3} - \sqrt{x + 3} \cdot \left[\sqrt{x + h + 3} + \sqrt{x + 3}\right] = x + h + 3 - \left(x + 3\right) = x + h + 3 - x - 3 = h$

f'(x)=lim_(h->0)(cancelx+h+cancel3-cancelx-cancel3)/(h(sqrt(x+h+3)+sqrt(x+3))

f'(x)=lim_(h->0)(cancelh)/(cancelh(sqrt(x+h+3)+sqrt(x+3))

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{1}{\sqrt{x + h + 3} + \sqrt{x + 3}}$

$f ' \left(x\right) = \frac{1}{\sqrt{x + 3} + \sqrt{x + 3}}$

$f ' \left(x\right) = \frac{1}{2 \sqrt{x + 3}}$